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2 years ago

How long would a sample of 26g of Nickel-63 take to decay down to

1 answer:

4 0

I looked up the half life for nickel-63 and then used the half life formula which is No=Nt*e^(kt) so the final answer is 20000 years for 26g of nickel-63 to decay to 6.5g

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Match the following terms with their definitions:

stepan [7]

Hi,

Homogeneous mixture 2
Heterogeneous 3
Mixture 1
Solution 5
Compound 4

6 0

3 years ago

Read 2 more answers

Each equation is incorrect. find the errors, then rewrite and balance each equation. cl2 + nai → nacl2 + i nh3 → n + h2 na + o2

VikaD [51]

1) Balanced chemical reaction: Cl₂ + 2NaI → 2NaCl + I₂.
Chlorine and iodine are diatomic molecules.
2) Balanced chemical reaction: 2NH₃ → N₂ + 3H₂.
Nitrogen and hydrogen are diatomic molecules.
3) Balanced chemical reaction: 4Na + O₂ → 2Na₂O.
Sodium in compounds has oxidation number +1 and oxygen -2.

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3 years ago

An atom of titanium has 22 protons and 26 neutrons. If the atoms loses one electron, what will be the charge on the ion that for

Alja [10]

It’s +21 (c) if i answered late my apologies

5 0

3 years ago

What is the mass of a bacterial cell?

SVEN [57.7K]

Thus, it was found that the intestinal microbiota has approximately 100 trillion bacterial cells - about ten times more than the total number of cells in the human body.

6 0

3 years ago

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Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The

qaws [65]

Answer:

$E_{ionization}=5.45\times 10^{-19}\ J$

Explanation:

$E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

So, $n_i=2$ and $n_f=\infty$ (As the hydrogen has to ionize)

Thus,

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2} - \dfrac{1}{{\infty}^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2})\ J$

$E_{ionization}=5.45\times 10^{-19}\ J$

4 0

3 years ago